Amrith Kumar added the comment:
As requested ...
>>> urlparse.urlparse('http://www.google.com:/abc')
ParseResult(scheme='http', netloc='www.google.com:', path='/abc', params='',
query='', fragment='')
I submit to you that this should generate an error.
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Python tracker <rep...@bugs.python.org>
<http://bugs.python.org/issue28841>
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